How to calculate Pr(Diseased | 2 Positive Tests)?

P(T) = P(T | D) + P (T | !D) = 0.98 + .05 = 1.03 ...?

That is incorrect. P(T) = P(T | D) x P(D) + P(T | !D) x P(!D) (remember Bayes' Rule?)

I think this alone is enough information for you to solve it. I must post this as a comment but due to no rep, am posting it as an answer. Hope it helps.

To get a handle on what Bayes’ rule is really doing in a simple case like your first problem, look at a little table:

$$\begin{array}{r|cc|c} &T&\lnot T&\text{Total}\\ \hline D&\color{red}{0.0098}&\color{red}{0.0002}&0.01\\ \lnot D&\color{red}{0.0495}&\color{red}{0.9405}&\color{blue}{0.99}\\ \hline \text{Total}&\color{green}{0.0593}&\color{green}{0.9407}&\color{blue}{1} \end{array}$$

What’s in black shows the basic setup: the number in row $D$ and column $T$ will be the probability that you have the disease and test positive, for instance. (I’ve used $\lnot D$, meaning $\text{not }D$, for the event of not having the disease.) I’ve also written in black the one number that you were explicitly given that goes in the table, namely, the probability that a randomly chosen person has the disease. Obviously the probability in the lower righthand corner must be $1$, and the probability that a randomly chosen person does not have the disease must be $1-0.01=0.99$, so I added those in blue.

Next, we know that $98$% of the $1$% who have the test will test positive, so the probability that a randomly chosen person both has the disease and tests positive is $0.98\cdot0.01=0.0098$, and the probability that such a person both has the disease and tests negative must be $0.01-0.0098=0.0002$; I’ve added these in red.

Similar reasoning tells us that the probability that a randomly chosen person does not have the disease and tests negative is $0.95\cdot0.99=0.9405$, so the probability that such a person does not have the disease but tests positive is $0.99-0.9405=0.0495$; I’ve added these in red as well.

At this point we can complete the table by adding the $T$ and $\lnot T$ columns to get the missing totals on the bottom line.

Now, suppose that you test positive. That means that you’re one of the $14.75$% represented by the first column of the table, and you want to know your chance of being one of the $0.98$% who test positive and have the disease. That probability is $\dfrac{0.0098}{0.0593}\approx 0.1653$: you have almost one chance in six of having the disease.

Bayes’ rule boils all of this down to a single formula. We ended up calculating the ratio $$\frac{\mathrm{Pr}(D\cap T)}{\mathrm{Pr}(T)}\;,$$ but we got the numerator by calculating $\mathrm{Pr}(T|D)\mathrm{Pr}(D)$, so in fact we calculated

$$\mathrm{Pr}(D|T)=\frac{\mathrm{Pr}(T|D)\mathrm{Pr}(D)}{\mathrm{Pr}(T)}\;,$$

which is Bayes’ rule.

In this case we also had to work a bit to get the denominator; that’s typical.